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Need closure (if not guidance) on this problem also:
An Atwood machine consists of two masses, [itex]m_1[/itex] and [itex]m_2[/itex], which are connected by a massless inelastic cord that passes over a pulley. If the puley has radius r and moment of inertia I about its axle, determine the acceleration of the masses [itex]m_1[/itex] and [itex]m_2[/itex].
The pulley has two forces acting on it which are providing the torque. These two forces are the tensions of the rope. Let [itex]T_1[/itex] and [itex]T_2[/itex] be the tensions caused by [itex]m_1[/itex] and [itex]m_2[/itex] respectively. Using the definition of torque, then
An Atwood machine consists of two masses, [itex]m_1[/itex] and [itex]m_2[/itex], which are connected by a massless inelastic cord that passes over a pulley. If the puley has radius r and moment of inertia I about its axle, determine the acceleration of the masses [itex]m_1[/itex] and [itex]m_2[/itex].
The pulley has two forces acting on it which are providing the torque. These two forces are the tensions of the rope. Let [itex]T_1[/itex] and [itex]T_2[/itex] be the tensions caused by [itex]m_1[/itex] and [itex]m_2[/itex] respectively. Using the definition of torque, then
[tex]
(T_1 + T_2)r = I\alpha
[/tex]
I'll assume that the angular acceleration of the cord around the pulley is the same as the angular acceleration of the pulley. Since the two masses and the cord have the same acceleration a (magnitude-wise), then I can use [itex]a = \alpha{r}[/itex]. Using F = ma on the two masses, I can solve for the tensions and therefore solve for a in the expression above. I get(T_1 + T_2)r = I\alpha
[/tex]
[tex]
a = \frac{g(m_1 + m_2)}{I/r^2 - m_1 + m_2}
[/tex]
a = \frac{g(m_1 + m_2)}{I/r^2 - m_1 + m_2}
[/tex]